# Spaceship launching instrument package

1. Feb 13, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in space at distance 5R from the center of the planet, the ship fires an instrument package with speed v0 (at angle $\theta$ with the line passing through the spaceship and the center of the planet). The package has mass m, which is much smaller than the mass of the spaceship. For what angle $\theta$ will the package just graze the surface of the planet ?

2. Relevant equations

3. The attempt at a solution

The origin of the frame is located at the center of the planet.
In the system 'spaceship + package', the external torque is 0 because the forces acting on it are radial. So the angular momentum is conserved. Furthermore, the final velocity vector of the package has to be perpendicular to the position vector, otherwise the instrument package would crash on the planet, so:

$L_i = 0 + 5mRv_0\sin(\theta) = 5mRv_0\sin(\theta)$
$L_f = 0 + mRv = mRv$

and then $\theta = \arcsin(\frac{v}{5v_0})$. But I feel it is not all because I'm not using the fact that m is a lot smaller than the mass of the spaceship ? Maybe $v$ has an expression but I don't see how to find it. Any idea ?

2. Feb 13, 2015

### BvU

You want $L_f$ big enough for the package not to fall on the ground...

The other thingy given is just so you don't worry about L for the spaceship and the package before and after shooting it off.

3. Feb 13, 2015

### geoffrey159

Hmm, am I wrong here ?

Ok, thanks!

4. Feb 13, 2015

### BvU

No, not wrong: what you write is correct, but as you state, it's not all. You want v big enough so the package doesn't fall on the ground. Where does the force that changes v come from ?

5. Feb 13, 2015

### geoffrey159

Also, we want no radial acceleration for the package. With Newton second law:

$m\frac{v^2}{R} = \frac{Gm}{R^2} (M +\frac{ M_{\text{s}}}{26}) \Rightarrow v = \frac{\sqrt{G}}{R} \sqrt{(M +\frac{ M_{\text{s}}}{26})}$

Is that correct ?

6. Feb 13, 2015

### BvU

If MS is the mass of the ship, perhaps you can ignore that in comparison with M of planet ?
Don't quite see where the $M_S\over 26$ came from. Can you elaborate ?

7. Feb 13, 2015

### geoffrey159

Yes, plus I made a mistake. When the package is at the surface of the planet, the spaceship attracts the package with the force :

$\vec F_s = - GmM_s \frac{5R\hat x + R\hat r}{|5R\hat x + R\hat r|^3}$

When the package just arrives at the surface of the planet $\hat x$ and $\hat r$ are orthogonal, so by the pythagorean theorem:
$|5R\hat x + R\hat r|^3 = ((5^2+1) R^2)^{3/2} = 26^{3/2} R^3$, and the radial componant of $\vec F_s$ is $F_{s,r} = - GmM_s \frac{R}{|5R\hat x + R\hat r|^3} = -\frac{GmM_s}{26^{3/2}R^2}$

8. Feb 13, 2015

### BvU

This is getting too complicated for me. We had a simple momentum conservation with a degree of freedom in $v$ at the surface. Gravitational potential (*) will tell you what $v$ is. And it has to be enough to make sure the trajectory curvature radius > planet radius. That's all, I would say.

(*) Gravitational potential due to the planet.

And the spaceship doesn't attract the package; that's why its mass doesn't feature in the exercise text.

9. Feb 13, 2015

### BvU

Coming back to this one: No. $v=\sqrt{GM\over R}\ \$ (factor R1/2)

10. Feb 13, 2015

### Staff: Mentor

If I may make a suggestion or two that might simplify the analysis?

In these sorts of problems where you're looking for a relationship (here an angle) and no specific values are given for the quantities involved other than ratios (staring radius is 5R where R is one Earth radius), it's convenient to choose a system of units where they're all equal to 1.

Then you can set u = GM = 1. You can also use what are known as "specific" quantities for angular momentum and energy. They are essentially the angular momentum and energy per unit mass of the object.

So the specific angular momentum would be h = r v sin(Θ), and the specific mechanical energy would be

$ξ = \frac{v^2}{2} - \frac{1}{r}$

Both of these quantities must remain constant when there are no external forces or torques at play. Write the expressions for both at the two instants of interest. You have the pertinent angles for the specific angular momentum in each case.

11. Feb 13, 2015

### geoffrey159

Ok I get it, if the spaceship does not attract the package, its what we get (sorry for my rusty algebra :-))

But Gneil has a point, if the package is acted by the planet's attraction, there is energy conservation and we should have:

$\frac{1}{2}mv^2 - \frac{GmM}{R} = \frac{1}{2}mv_0^2 - \frac{GmM}{5R}$

and then $v = \sqrt{v_0^2 + \frac{8}{5}\frac{GM}{R}}$

Why isn't it the same v then ?

12. Feb 13, 2015

### BvU

It's not the same v because the package falls towards the planet and gains kinetic energy (from losing potential energy). The game is to choose theta in such a way that it 'just misses the planet'. Like the Newton's[/PLAIN] [Broken] cannonball that is fired at 8000 m/s (number C) and 'falls' around the earth.

Only here it's fired at high altitude (and lower speed, and at a different planet,I suppose :) )

You can fire the cannon yourself here (sorry for getting carried away )

Last edited by a moderator: May 7, 2017